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4

No, one cannot obtain a solution of this PDE by adding a solution of heat equation to a solution of the wave equation. (With linear PDE, we can combine solutions of the same equation to make new ones; but your situation is different). Your PDE is known as damped wave equation and is solved here. It does inherit some features from the heat and wave equation. ...


3

As you pointed out we have \begin{align*} -\Delta w & = 0 \text{ in }\Omega \\ w & = 0 \text{ on }\partial\Omega \end{align*} and \begin{align*} -\Delta (-w) & = 0\text{ in }\Omega \\ -w & = 0\text{ on }\partial\Omega \end{align*} Thus we can use the maximum principle for $w$ and $-w$, yielding $\max_\Omega w = \max_{\partial\Omega} w =0$ and ...


2

Here are three (somewhat standard) references: Jurgen Jost, "Riemannian Geometry and Geometric Analysis." Peter Li, "Lecture Notes on Geometric Analysis." Thierry Aubin, "Nonlinear Analysis on Manifolds. Monge-Ampere Equations." Jost's book is on its sixth edition. Aubin's book has a first and second edition, although my understanding is that the first ...


2

You are correct to use separation of variables, the next step is to solve the two independent differential equations. First, I believe that you should use $ \frac {X''}{X} = \frac{4T'}{T} = - \lambda^2$ as this gives you a negative only answer. This will give you the following, $X(x) = A\sin( \lambda x)+B\cos( \lambda x)$ $T(t) = Ce^{\frac {-\lambda^2 ...


2

I'm guessing you mean $u\in C^2(\Omega)\cap C^0(\bar\Omega)$. $u$ is a continuous function on $\bar\Omega$, which is compact, hence reaches its extrema. Let $x_0$ be a maximum point. If $x_0\in\Omega$, then $Du(x_0)=0$ and $\Delta u(x_0)\leq 0$. By plugging this into your equation, we get $$ \underbrace{\Delta u(x_0)}_{\leq 0} + ...


2

$$ u(x,t) = V(x)\mathrm{e}^{st +ikx} $$ and $$ u_t =u_{xx} + \mu u $$ we find $$ su = \left(V''\mathrm{e}^{st +ikx} +2ikV'\mathrm{e}^{st +ikx}-k^2u\right) + \mu u $$ thus $$ \left(V'' +2ikV'\right)\mathrm{e}^{st +ikx}+(\mu -k^2-s)u = 0 $$ This is what I would of done without the answer you have shown, make the terms in the bracket zero. $$ \mu -k^2-s = 0 ...


2

Hint: $\left(\dfrac{\partial u}{\partial x}\right)^2-\left(\dfrac{\partial u}{\partial y}\right)^2+\dfrac{1}{2}\dfrac{(y^2-x^2)^2}{xy(x^2+y^2)}\dfrac{\partial u}{\partial x}\dfrac{\partial u}{\partial y}=0$ $2xy(x^2+y^2)\left(\dfrac{\partial u}{\partial x}\right)^2+(x^2-y^2)^2\dfrac{\partial u}{\partial x}\dfrac{\partial u}{\partial ...


2

It depends... the Laplacian is always defined with respect to some metric: $\Delta f = \nabla \cdot \nabla f$ and divergence requires a metric. Alternatively, you can define $\Delta$ as the gradient, in the sense of the calculus of variations, of the Dirichlet energy $\int \langle \nabla f, \nabla f\rangle\,dV$ and here again the metric is seen. My guess is ...


2

The change $$ u(t,x)=e^{ax+bt}\,v(t,x) $$ for appropriate choice of $a,b\in\mathbb{R}$ will transform the equation into $$ v_t=v_{xx},\quad v(0,x)=e^{-ax}\,g(x). $$


2

The full procedure is to set both sides equal to an arbitrary constant $\lambda$. Then we have to solve $f''(x) = \lambda f(x)$ with the boundary conditions $f(0)=f(1)=0$, and there are three cases: $\lambda > 0$ gives the hyperbolic case, and (as you can check for yourself) the only function in this family of solutions which satisfies the boundary ...


2

Just plugin the definition of $T$ and $Q_\sigma$. We have $$ u = \sigma Tu \iff Tu = \frac{u}{\sigma} $$ that is, by definition of $T$ iff $v := \frac{u}\sigma$ is the (unique) solution of $$ a^{ij}(x,u,Du)D_{ij}v + b(x,u,Du) = 0, \quad v|_{\partial \Omega} = \phi $$ Let here $v = \frac u\sigma$, this gives $$ \frac 1\sigma a^{ij}(x,u,Du) D_{ij}u + ...


2

Answer for part (a) should be $$\frac{1-ik}{1+k^2}$$ So by the inversion formula $$f(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}\frac{1-ik}{1+k^2} e^{ikx}\,dk$$ Now separate the integrand into real part and imaginary part. Notice that $f(x)$ is a real function, so the imaginary part in the integrand should be zero. The real part is exactly the integral in ...


1

Because $(\Delta f,f) \le -\lambda_1\|f\|^{2}$ for $f\in\mathcal{D}(-\Delta)$, then \begin{align} \frac{d}{dt}\|T(t)f\|^{2}& =(\Delta T(t)f,T(t)f)+(T(t)f,\Delta T(t)f) \\ & \le -2\lambda_1(T(t)f,T(t)f)= -2\lambda_1\|T(t)f\|^{2},\;\; f \in\mathcal{D}(-\Delta) \end{align} Hence, $$ \frac{d}{dt}\left( e^{2\lambda_1 ...


1

Let's put the tutorial solution into a normalized form. Which means either using only hyperbolic functions or expressing everything exponential or hyperbolic in $e^x$ and $e^{-x}$. Aiming for hyperbolic functions, one gets, using $e^x=\cosh(x)+\sinh(x)$, \begin{align} u_e(x)&=\frac{3e+1}{\cosh(1)}\sinh(x)-3e^x+x-x^2-2 \\ \\ ...


1

First: The definition of conformality looks odd; did you mean something like $Dg(x)^{t} \cdot Dg(x) = C(x) I$, i.e., $$ \left\langle Dg(x)u, Dg(x)v \right\rangle = C(x) \langle u, v\rangle \quad\text{for all $u$, $v$} $$ instead? Second: Inside the parentheses, note that $x^{t}x = \|x\|^{2}$.


1

Solve $$ v_t-v_{xx}=0,\quad v(0,t)=v(\ell,t)=0,\quad v(x,0)=f(x) $$ and $$ w_t-w_{xx}=g(x,t),\quad w(0,t)=w(\ell,t)=0,\quad w(x,0)=0. $$ The $u=v+w$. Standard separation of variables gives $v$ in the form $$ v(x,t)=\sum_{n=1}^\infty a_n\,e^{-\bigl(\tfrac{k\,\pi}{\ell}\bigr)^2\,t}\,\sin\frac{k\,\pi\,x}{\ell}. $$ To find $w$ develop $g$ in a Fourier series ...


1

It looks like you have a slight mistake, in that \begin{equation} \max_{\Omega}(-(u_1 - u_2)) = - \min_{\Omega}(u_1 - u_2) \end{equation} not $+\min_{\Omega}(u_1 - u_2)$ which you have written. This means that \begin{equation} \min_{\Omega}(u_1 - u_2) = -\max_{\partial\Omega}(-(g_1 - g_2)) = \min_{\partial \Omega}(g_1 - g_2). \end{equation} You already ...


1

a more general solution to $u_x + u_y = 1$ is $u(x,y)= Ax + (1-A)y+C$ the use of the variable $y$ in the expression of the boundary condition is unfortunate , we could say $u(p, \frac{p}{2}) = p$ for all real numbers $p$. plug that into the general solution, keeping in mind that this is an identity in $p$, by equating coefficients you should get that $C=0$ ...


1

The Poisson kernel for balls with centre $0$ in $\mathbb{R}^n$ is given by $$P(x,\zeta) = \frac{1}{\lVert\zeta\rVert\omega_{n-1}} \cdot \frac{\lVert\zeta\rVert^2 - \lVert x\rVert^2}{\lVert \zeta - x\rVert^n},$$ and for any function $u$ harmonic on the open ball with radius $R$ and continuous on the closed ball, in particular for entire harmonic functions, ...


1

For $H^1$- functions, this does not hold in general. It is basically the same argument that $L^2$- functions generally do not vanish at $\infty$, you just have to take such a function and integrate it, for example a bump function where the bumps get thinner when you go outside. For $H^2$-functions however, it does hold: Take $$\int_0^a ...


1

Use a sequence of harmonics with increasing frequency and decreasing amplitude: $$f_n(x)= A_n \sin \left(\frac{\pi n}{L} x\right)$$ As long as $A_n\to 0$, the values at time $T$ tend to zero in the $L^p$ sense. On the other hand, the solution at time $t<T$ is $$f_n(x)= A_n e^{\lambda (\pi n/L)^2 (T-t)}\sin \left(\frac{\pi n}{L} x\right)$$ which, for ...


1

For every $x\in \overline{\Omega'}$, we have $$\sup \{ u(z) : z \in B_r(x)\} \leqslant 3^n\inf \{ u(z) : z \in B_r(x)\}\tag{1}$$ by assumption. From the open cover $\{ B_r(x) : x \in \overline{\Omega'}\}$ of $\Omega'$ we extract a finite subcover $\{ B_m : 1 \leqslant m \leqslant k\}$. If $B_m \cap B_j \neq \varnothing$, then we have $$\sup \{ u(z) : z ...


1

$$ F: \big(u(t),\lambda\big) \to \frac{d^2}{dt^2}u + \lambda \sin u $$ Expand $\sin u$ into series around $u=0$, get linearization of $F$ operator: $$ F\big(u,\lambda\big) = \frac{d^2}{dt^2}u + \lambda \left( u - \frac{u^3}{3!} + \cdots\right) \approx \bigg[ \frac{d^2}{dt^2} + \lambda I\bigg] u, $$ where $I$ is an identity mapping. Denote $L:X\times R\to ...


1

Note that, with the fact that $a \neq b$, $$\frac{\log(b)}{\log(\frac{b}{a})} - \frac{\log(a)}{\log(\frac{b}{a})} = \frac{\log(b) - \log(a)}{\log(b) - \log(a)} = 1$$ This tells that $$\frac{n\pi\log(b)}{\log(\frac{b}{a})} = n\pi +\frac{n\pi\log(a)}{\log(\frac{b}{a})}$$ and that $$\cos(\frac{n\pi\log(b)}{\log(\frac{b}{a})}) = ...


1

No, the Laplacian does not map $L^2$ to $L^2$. But one can consider it as an unbounded operator from a dense subspace of $L^2$ into $L^2$. This dense subset includes $C_0^\infty$, since the Laplacian maps $C_0^\infty$ into $C_0^\infty$. To discuss self-adjointness, one must impose boundary conditions and close the operator. I suggest reading a book about ...


1

You need to additionally assume that $S(t)$ is a symmetric matrix for all $t$, and that it has even dimension $2k \times 2k$. Here are some hints (without giving away the whole solution). Let’s first simplify notation (to avoid using the $t$ variable in two different ways, which is confusing to me). Define: $$H(u, z) = (1/2)z^T S(u) z$$ Thus: ...


1

Here's a summary of, and extending remarks and explanations to, the comments. Note that my description of $u(t,x)$ is based on experimental observations, not formal proofs. I plotted the sum for $u(t,x)$ for a few values of $t$: within a moderate range for $x$, it would converge, although I did need extended accuracy (used Maple for this). My understanding ...


1

$$ |y-x|^2=|y|^2-2\,y\cdot x+|x|^2=1-2\,y\cdot x+|x|^2. $$ $$ |x|^2\,\Bigl|\,y-\frac{x}{|x|^2}\Bigr|^2=|x|^2\Bigl(|y|^2-\frac{2\,y\cdot x}{|x|^2}+\frac{1}{|x|^2}\Bigr)=|x|^2-2\,y\cdot x+1. $$



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